Do we have to take the absolute value of the jacobian ONLY if it is a number? if we want to evaluate the integration
$$I=\int\int(x^3y^3)(x^2+y^2)dA$$
over the region bounded by the curves
$$xy=1,\\xy=3,\\x^2-y^2=1,\\x^2-y^2=4$$
I used the transformation 
$$u=xy,\\v=x^2-y^2$$
I found that the jacobian will be 
$$J=\frac{1}{-2y^2-2x^2}$$
Do I have to get the absolute value of the jacobian?
If I did not take the absolute value , I will get the result of the integration = - 30
if I take the absolute value
$$J=\frac{1}{2y^2+2x^2}$$
I will get the result= + 30 .
My friend told me we take absolute value of the jacobian only if it is a number .. if this is right .. why we do not take the absolute value if the jacobian is a function?..I think we are sure here that the jacobian is negative since we have x and y squared , so we have to take the absolute value!
Another question, if we have to take always the absolute value of the jacobian (whether it is a number or function)  :
if the jacobian is for example 
$$J=-2x+y$$
It will be positive for some values of x and y only ! .. how can we apply the absolute value inside the double integration? 
 A: If the Jacobian is negative, then the orientation of the region of integration gets flipped.
You have to take the absolute value ALWAYS.
A: I'm adding an additional answer, in case the issue is actually with the integration of an absolute value, recall the definition of the absolute value: 
$\begin{eqnarray*}
\left|-2x+y\right| \quad & = & \quad 
     \begin{cases}
       -2x+y, & \text{if } -2x+y \geq 0, \quad \text{ (i.e. if the quantity is positive)}\\
       -(-2x+y), & \text{if } -2x+y < 0, \quad \text{ (i.e. if the quantity is negative)}
     \end{cases} \\[8pt]
     %
 \quad & = & \quad 
     \begin{cases}
     -2x +y, & y \geq 2x   ; \\
     2x - y, & y < 2x   
   \end{cases}
\end{eqnarray*}$
As such, if the domain of integration is in just one of these two halves of the plane, you can just apply the appropriate form of the absolute value; and if the domain of integration is a region containing a part of the line $y = 2x$, you simply break up the integral into multiple parts, where one of the two definitions applies. 
N.B. here I have expressed $y = y(x)$, but if you wanted to integrate with respect to $x$ first, you could simply rearrange the expression to have $x = x(y)$. 
